# If f(x) =sec^3(x/2)  and g(x) = sqrt(2x-1 , what is f'(g(x)) ?

Jan 21, 2016

#### Answer:

$f ' \left(g \left(x\right)\right) = \frac{3}{2} {\sec}^{3} \left(\frac{\sqrt{2 x - 1}}{2}\right) \tan \left(\frac{\sqrt{2 x - 1}}{2}\right)$

#### Explanation:

First, we should just find $f ' \left(x\right)$ so that we can then plug $g \left(x\right)$ into it.

To find $f ' \left(x\right)$, the primary and overriding issue is the third power. According to the chain rule, $\frac{d}{\mathrm{dx}} \left[{u}^{3}\right] = 3 {u}^{2} \cdot u '$, hence

$f ' \left(x\right) = 3 {\sec}^{2} \left(\frac{x}{2}\right) \cdot \frac{d}{\mathrm{dx}} \left[\sec \left(\frac{x}{2}\right)\right]$

Now, to differentiate the secant function, recall that (through the chain rule) $\frac{d}{\mathrm{dx}} \left[\sec \left(u\right)\right] = \sec \left(u\right) \tan \left(u\right) \cdot u '$, hence

$f ' \left(x\right) = 3 {\sec}^{2} \left(\frac{x}{2}\right) \cdot \sec \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right) \cdot \frac{d}{\mathrm{dx}} \left[\frac{x}{2}\right]$

Note that $\frac{d}{\mathrm{dx}} \left[\frac{x}{2}\right] = \frac{d}{\mathrm{dx}} \left[\frac{1}{2} x\right] = \frac{1}{2}$. Also, recognize that the secant functions can be multiplied with one another, yielding a simplified $f ' \left(x\right)$:

$f ' \left(x\right) = \frac{3}{2} {\sec}^{3} \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)$

To find $f ' \left(g \left(x\right)\right)$, simply plug in $\sqrt{2 x - 1}$ wherever there's an $x$ in $f ' \left(x\right)$.

$f ' \left(g \left(x\right)\right) = \frac{3}{2} {\sec}^{3} \left(\frac{\sqrt{2 x - 1}}{2}\right) \tan \left(\frac{\sqrt{2 x - 1}}{2}\right)$

Jan 21, 2016

#### Answer:

$f ' \left(g \left(x\right)\right) = \frac{3}{2 \left(\sqrt{2 x - 1}\right)} {\sec}^{3} \left(\frac{\sqrt{2 x - 1}}{2}\right) \cdot \tan \left(\frac{\sqrt{2 x - 1}}{2}\right)$

#### Explanation:

If the question meaning is $f ' \left(g \left(x\right)\right)$ as $\left(f \circ g\right) '$ then:

find $f \circ g = f \left(g \left(x\right)\right)$

$y = f \left(g \left(x\right)\right) = {\left[\sec \left(\frac{\sqrt{2 x - 1}}{2}\right)\right]}^{3}$

$g \left(x\right)$ is a composition too, therefore you can think:

$y = {u}^{3}$

$u = \sec \left(v\right)$

$v = \frac{\sqrt{z}}{2}$

$z = \left(2 x - 1\right)$

the Chain Rule tells

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dz}} \frac{\mathrm{dz}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{du}} = 3 {u}^{2} = 3 {\sec}^{2} \left(v\right) = 3 {\sec}^{2} \left(\frac{\sqrt{z}}{2}\right) = 3 {\sec}^{2} \left(\frac{\sqrt{2 x - 1}}{2}\right)$

$\frac{\mathrm{du}}{\mathrm{dv}} = \sec ' \left(v\right) = \tan \left(v\right) \cdot \sec \left(v\right) = \tan \left(\frac{\sqrt{z}}{2}\right) \cdot \sec \left(\frac{\sqrt{z}}{2}\right) =$
$= \tan \left(\frac{\sqrt{2 x - 1}}{2}\right) \cdot \sec \left(\frac{\sqrt{2 x - 1}}{2}\right)$

$\frac{\mathrm{dv}}{\mathrm{dz}} = \frac{1}{2} \cdot \left(\frac{1}{2 \sqrt{z}}\right) = \frac{1}{4 \sqrt{2 x - 1}}$

$\frac{\mathrm{dz}}{\mathrm{dx}} = 2 \cdot 1 + 0$

Thus:

$y ' = 3 {\sec}^{2} \left(\frac{\sqrt{2 x - 1}}{2}\right) \cdot \tan \left(\frac{\sqrt{2 x - 1}}{2}\right) \cdot \sec \left(\frac{\sqrt{2 x - 1}}{2}\right) \cdot \frac{1}{{\cancel{4}}^{2} \sqrt{2 x - 1}} \cdot \cancel{2} =$
$= \frac{3}{2 \left(\sqrt{2 x - 1}\right)} {\sec}^{2} \left(\frac{\sqrt{2 x - 1}}{2}\right) \cdot \tan \left(\frac{\sqrt{2 x - 1}}{2}\right) \cdot \sec \left(\frac{\sqrt{2 x - 1}}{2}\right)$

Alternatively, once you get practice, more simply:

Given:

$h \left(x\right) = f \left(g \left(x\right)\right)$

$h ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

if $g \left(x\right) = g \left(i \left(x\right)\right) \implies g ' \left(x\right) = g ' \left(i \left(x\right)\right) \cdot i ' \left(x\right)$

thus you can apply the rule ricorsively.